Everything over $\Bbb{C}$. Say we have a smooth curve $C$ of genus $10$ which is a double cover of a smooth plane cubic curve. Therefore $C$ admits a 1-dimensional family of pencils of degree 4 (arising from the involutions on the cubic).

Can we deduce from this that $C$ does not admit any pencil of degree $3$ ? (in other words that $W^1_3( C )$ is empty)

  • 1
    $\begingroup$ at least for base point free pencils, how's this? If C admits maps of degrees both 3 and 4 to P^1, then the product map to P^1xP^1 is generically injective, so the genus of C is ≤ 6. The same argument (of Castelnuovo) shows a curve with base point free pencils of relatively prime degrees d,e has genus ≤ (d-1)(e-1). $\endgroup$ – roy smith Jan 9 '16 at 18:00
  • $\begingroup$ technically we need also to rule out a hyperelliptic curve, since then there would be g(1,3)'s with base point. But among the curve of g(1,4)'s there is one whose general divisor does not contain a divisor of the g(1,2) so we get again that the genus is ≤ 3. $\endgroup$ – roy smith Jan 11 '16 at 23:57

For every integer $p_a>4$, there does not exist a smooth, projective, geometrically connected curve of genus $p_a$ that admits both a degree $2$, finite, flat morphism, $f:C\to E$, to a smooth plane cubic $E$ and a degree $3$, finite, flat morphism, $g:C\to \mathbb{P}^1$, to the projective line. Probably this can be proved directly from the geometric Riemann-Roch theorem, but the argument below just uses the adjunction formula. Note also, a complete intersection in $\mathbb{P}^2\times \mathbb{P}^1$ of hypersurfaces of bidegrees $(3,0)$ and $(1,2)$ is a curve of arithmetic genus $p_a=4$ that does admit a degree $2$ morphism to $E$ and a degree $3$ morphism to $\mathbb{P}^1$. So the inequality $p_a>4$ is necessary.

By way of contradiction, assume that there is such a curve. Consider the product morphism, $$(f,g):C\to E\times\mathbb{P}^1.$$ Denote the image Cartier divisor by $$i:B\to E\times \mathbb{P}^1.$$ This is a closed, integral (possibly singular) curve in $E\times \mathbb{P}^1$. Denote by $$h:C\to B,$$ the unique finite, surjective morphism such that $i\circ h$ equals $(f,g)$.

What is the degree of $h$? For every invertible $\mathcal{O}_E$-module $\mathcal{A}$ of degree $1$, the invertible $\mathcal{O}_C$-module $h^*(i^*\text{pr}_E^*\mathcal{A})$ has degree $2$. Thus, the degree of $h$ divides $2$. Similarly, the degree of $h^*(i^*\text{pr}_{\mathbb{P}^1}^*\mathcal{O}(1))$ has degree $3$. Thus, the degree of $h$ also divides $3$. Therefore the degree of $h$ equals $1$, i.e., $h$ is a birational equivalence. In particular, the arithmetic genus of $B$ is at least as large as the geometric genus $2p_a-2 > 6$ of $C$.

Because $\text{Pic}(E\times \mathbb{P}^1)$ equals $\text{Pic}(E)\times \mathbb{Z}$, there is an isomorphism of invertible sheaves $$\mathcal{O}_{E\times \mathbb{P}^1}(\underline{B}) \cong \text{pr}_E^*\mathcal{L}\otimes \text{pr}_{\mathbb{P}^1}^*\mathcal{O}(d),$$ for an invertible sheaf $\mathcal{L}$ on $E$ and for an integer $d$. By the computations above, $d$ must equal $2$ and $\mathcal{L}$ must have degree $3$. By the adjunction formula, there is an isomorphism of invertible sheaves, $$\omega_B \cong \left( \text{pr}_E^*\mathcal{L} \right)|_B.$$ The degree of this invertible sheaf on $B$ equals $6$, which is strictly less than $2p_a-2$.

Edit. Let $E$ and $F$ be smooth projective curves such that the natural map $\text{Pic}(E)\times \text{Pic}(F) \to \text{Pic}(E\times F)$ is an isomorphism. Let $C$ be a smooth, projective curve. Let $\epsilon:C\to E$ and $\phi:C\to F$ be finite, flat morphisms such that the product morphism $(\epsilon,\phi):C\to E\times F$ is birational to its image, e.g., this holds if $\text{deg}(\epsilon)$ and $\text{deg}(\phi)$ are relatively prime. Then the same argument as above implies the following inequality, $$p_a(C) \leq (\text{deg}(\epsilon)-1)(\text{deg}(\phi)-1) + p_a(E)\text{deg}(\epsilon) + p_a(F)\text{deg}(\phi). $$ In particular, when $F$ is $\mathbb{P}^1$ and $d$ denotes $\text{deg}(\phi)$, so that we are considering $g^1_d$s on $C$, this reduces to the following inequality, $$d \geq \frac{(2p_a(C)-2) - \text{deg}(\epsilon)(2p_a(E)-2)}{2(\text{deg}(\epsilon)-1)} = \frac{\text{deg}(\text{Branch}(\epsilon))}{2(\text{deg}(\epsilon)-1)}.$$ Finally, if also $E$ has genus $1$, this becomes the following, $$d \geq \frac{p_a(C)-1}{\text{deg}(\epsilon)-1}.$$

Second Edit. Thanks to Felipe Voloch who recognized this inequality. It is the Castelnuovo-Severi inequality.

  • $\begingroup$ Really great answer. Thank you Jason! $\endgroup$ – Heitor Dec 19 '15 at 11:56
  • $\begingroup$ Only, I do not understand the very last formula. Applied to the above example gives $d\geq9$ which is not the case, right? $\endgroup$ – Heitor Dec 19 '15 at 11:59
  • $\begingroup$ @Heitor: "Applied to the above example gives $d\geq 9$ which is not the case, right?" The degree of $\epsilon$ equals $2$, and the degree $d$ of $\phi$ equals $3$. Thus the inequality is $3 \geq (p_a(C)-1)/(2-1)$, i.e., $p_a(C) \leq 4$. So this is the same inequality as in the main part of the argument. $\endgroup$ – Jason Starr Dec 19 '15 at 12:38
  • $\begingroup$ Oh yes, right. My confusion :) Thank you $\endgroup$ – Heitor Dec 19 '15 at 12:41
  • 1
    $\begingroup$ AKA Castelnuovo-Severi inequality. $\endgroup$ – Felipe Voloch Dec 19 '15 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.